# Chapter 3 Second Order Linear Equations

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1 Partial Differential Equations (Math 3303) A Õæ Aë ú GA É Chapter 3 Second Order Linear Equations Second-order partial differential equations for an known function u(x, y) has the form F (x, y, u, u x, u y, u xx, u xy, u yy ) = 0. A second-order PDE is linear if it can be written as A(x, y)u xx + B(x, y)u xy + C(x, y)u yy + D(x, y)u x + E(x, y)u y + F (x, y)u = G(x, y). (1) In this chapter we will study equation (1). We assume that A, B, and C are functions with continuous second-order derivatives in a domain Ω R 2 and they do not vanish simultaneously in Ω. 3.1 Classification of Linear PDEs of Second Order Equation (1) can be classified into three distinct types: hyperbolic (e.g., the wave equation), parabolic (e.g., the heat equation), and elliptic equations (e.g., the Laplace equation). It turns out that solutions of equations of the same type share many exclusive qualitative properties. We will show that by a certain change of variables any equation of a particular type can be transformed into a canonical form which is associated with its type. Definition. (Hyperbolic, Parabolic, and Elliptic PDE) Consider the second-order linear equation (1). (1) We define the discriminant of equation (1) to be (x, y) = B 2 4AC. (a) If (x, y) > 0 at a point (x, y), then equation (1) is said to be hyperbolic at (x, y). (b) If (x, y) < 0 at a point (x, y), then equation (1) is said to be elliptic at (x, y). (c) If (x, y) = 0 at a point (x, y), then equation (1) is said to be parabolic at (x, y). (2) Equation (1) is said to be hyperbolic, elliptic, or parabolic in a domain Ω R 2 if it is, respectively, hyperbolic, elliptic, or parabolic at every point of Ω. Example. Classify the following PDEs as hyperbolic, elliptic, or parabolic. (1) u xx + 2u xy + u yy u x + xu y = 0. (2) x 2 u xx y 2 u yy = xy. (3) u xx + u xy + 2u yy = u x 3yu. 1

2 Canonical forms of second-order linear equations We can use the change of variables to transform equation (1) into simpler forms. ξ = ξ(x, y), η = η(x, y) Lemma. The type of a linear second-order PDE is invariant under a nonsingular transformation ξ = ξ(x, y), η = η(x, y). Proof. Definition. (Canonical forms of linear second-order equations) (1) The canonical form of a hyperbolic PDE is (2) The canonical form of an elliptic PDE is (3) The canonical form of a parabolic PDE is u ξη = H(ξ, η, u, u ξ, u η ). u ξξ + u ηη = H(ξ, η, u, u ξ, u η ). u ξξ = H(ξ, η, u, u ξ, u η ). Theorem 1. (Canonical form of hyperbolic equations) Suppose that equation (1) is hyperbolic in a domain Ω. There exists a coordinate system (ξ, η) in which the equation has the canonical form Proof. u ξη = H(ξ, η, u, u ξ, u η ). Remark. In order to find the transformation ξ, η, we have to solve a PDE φ x λφ y = 0. We solve this PDE by solving the ODE dy = λ and obtaining φ(x, y) = c. dx Example. Show that the partial differential equation 2u xx + 3u xy + u yy + xu x + u = 0 is hyperbolic and find an equivalent partial differential equation in canonical form. Theorem 2. (Canonical form of parabolic equations) Suppose that equation (1) is parabolic in a domain Ω. There exists a coordinate system (ξ, η) in which the equation has the canonical form Proof. u ξξ = H(ξ, η, u, u ξ, u η ). 2

3 Remark. In the above theorem we choose η(x, y) as a solution of the linear first-order PDEs and we may choose ξ(x, y) = x. η x + B 2A η y = 0, Example. Show that the partial differential equation x 2 u xx 2xyu xy + y 2 u yy + xu x = x 2 + u y is a parabolic equation and find its canonical form. Theorem 3. (Canonical form of elliptic equations) Suppose that equation (1) is elliptic in a domain Ω. There exists a coordinate system (ξ, η) in which the equation has the canonical form Proof. u ξξ + u ηη = H(ξ, η, u, u ξ, u η ). Example. Show that the partial differential equation u xx + x 2 u yy = yu y is elliptic in the region x 0 and find an equivalent canonical form. Characteristic curves In order to transform a second-order linear equation A(x, y)u xx + B(x, y)u xy + C(x, y)u yy + D(x, y)u x + E(x, y)u y + F (x, y)u = G(x, y), (2) into canonical form we have to find a nonsingular transformation In the three cases we have to solve where λ 1,2 are the roots of the equation ξ = ξ(x, y), dy dx = λ 1,2, η = η(x, y). Aλ 2 + Bλ + C = 0. Definition. (Characteristic curves) The graph of the function y = y(x) is called a characteristic curve for the second-order linear partial differential equation (2) if it satisfies dy dx = λ 1,2, where λ 1,2 are the roots of the equation Aλ 2 Bλ + C = 0. Example. Find the characteristic curves for the partial differential equation 2u xx + 3u xy + u yy + xu x + u = 0. Remark. Second-order partial differential equations in more than two independent variables can also be classified into types, including parabolic, elliptic, and hyperbolic. However, it is not usually possible to reduce such equations to simple canonical forms. 3

4 Exercises (1) Classify the each of following PDEs as hyperbolic, parabolic, or elliptic and find an equivalent PDE in canonical form. (a) u xx + 2u xy + u yy = u x xu y (b) u xx + 2u xy + 5u yy = 3u x yu (c) 3u xx + 10u xy + 3u yy = 0 (d) u xx + 6u xy + u yy 4uu x = 0 (e) 4u xx 8u xy + 4u yy = 1 (2) Determine where the given PDE is hyperbolic, parabolic, or elliptic. (a) u xx + 2yu xy + 5u yy = 15x + 2y (b) x 2 u xx + 4y u yy u = 0 (c) x 2 y u xx + xy u xy y 2 u yy = 0 (d) xy u xx x u xy + u yy uu x = 3 (e) sin x u xx + 2 cos x u xy + sin x u yy = 0 (f) x ln x u xx + 4u yy u x + 3xyu = 0 (g) u xx + x u xy + y u yy = 0 (h) (1 x 2 ) u xx 2xy u xy + (1 y 2 ) u yy + xu x + 3x 2 yu y 2u = 0 (3) Consider the Tricomi PDE y u xx + u yy = 0. (a) Show that the equation is hyperbolic when y < 0, parabolic when y = 0, and elliptic when y > 0. (b) Find an equivalent PDE in canonical from when y < 0. (c) Find an equivalent PDE in canonical from when y = 0. (d) Find an equivalent PDE in canonical from when y > 0. (4) Find regions in which x 2 u xx + 4u yy = u hyperbolic, parabolic, and elliptic. In each region, find an equivalent PDE in canonical form. (5) Show that y 2 u xx 2xy u xy + x 2 u yy = 0 is everywhere parabolic. Find an equivalent PDE in canonical form valid in regions not containing points on the x axis. (6) Show that u xx + x 2 u xy ( 1 2 x2 + 1 ) u yy = 0 4 is hyperbolic in the entire xy plane and find its canonical form (7) Show that x u xy + y u yy = 0 is hyperbolic when x 0 and find its canonical form 4

5 (8) Consider A(x, y)u xx + B(x, y)u xy + C(x, y)u yy + D(x, y)u x + E(x, y)u y + F (x, y)u = G(x, y), A, B, C, D, E, F are constants. (a) Show that if the equation is hyperbolic, then its canonical form is where a, b, c are constants. u ξη + au ξ + bu η + cu = H(ξ, η), (b) Show that if the equation is parabolic, then its canonical form is where a, b, c are constants. u ξξ + au ξ + bu η + cu = H(ξ, η), (c) Show that if the equation is elliptic, then its canonical form is where a, b, c are constants. u ξξ + u ηη + au ξ + bu η + cu = H(ξ, η), (d) Show that the first-order derivative terms u ξ and u η in (a), (b), and (c) can be eliminated by a change of dependent variable for appropriate constants p and q. w(ξ, η) = e pξ+qη u(ξ, η) (9) Use the results of Exercise (8) to find a simplified canonical form for each of the following PDEs: (a) u xx + 2u xy + 5u yy 3u x = 0. (b) u xx + 2u xy + u yy u x + u y = 0. (b) u xx + 6u xy + u yy 4u x = General Solution of Second-order Linear PDEs In this section we will study methods for solving second-order linear partial differential equations: A(x, y)u xx + B(x, y)u xy + C(x, y)u yy + D(x, y)u x + E(x, y)u y + F (x, y)u = G(x, y). (3) A simple special case of equation (3) occurs when A, B, C, D, E, F are constants. In such case equation (3) is called a second-order linear PDE with constant coefficients. Definition. (General solution) Consider the second-order linear PDE L(u) = A(x, y)u xx + B(x, y)u xy + C(x, y)u yy + D(x, y)u x + E(x, y)u y + F (x, y)u = G(x, y). (a) The general solution of the homogenous second-order linear PDE L(u) = 0 is the solution containing two arbitrary functions of one variable. 5

6 (b) If u h is the general solution of homogenous PDE L(u) = 0 and u p is any particular solution of the non-homogenous PDE L(u) = G(x, y), then u = u h + u p is called the general solution of non-homogenous PDE L(u) = G(x, y). Theorem 4. (Superposition Principle for Homogeneous Equation) If u 1,..., u k are solutions to a homogeneous linear equation L(u) = 0, then the linear combination, or superposition, u = c 1 u c k u k is a solution for any choice of constants c 1,..., c k. Proof. By linearity of L. Theorem 5. (Superposition Principle for Nonhomogeneous Equation) If u 1,..., u k are solutions to the nonhomogeneous linear systems L(u j ) = p j, j = 1,..., k, involving the same linear operator L. Then, for any constants c 1,..., c k, the linear combination, or superposition, u = c 1 u c k u k is a solution to nonhomogeneous equation L(u) = c 1 p c k p k. Proof. By linearity of L. Theorem 4 and Theorem 5 give us powerful tools for solving linear partial differential equations, which we will repeatedly use throughout this course. General solution by direct integration Some PDEs can be solved by direct integrations. Example 1. Find the solution of the PDE u xy = 6xy 2. Example 2. Solve the PDE subject to the conditions u yy = e y u y (x, 0) = x 2, u(x, 0) = e x. General solution by reduction to ODE If a give PDE contains partial derivatives with respect to only one independent variable, then it can be solved using methods of ordinary differential equations. Example 1. Find the general solution of the PDE u yy 4u y + 3u = 0. 6

7 Example 2. Solve the PDE xu xy + 2u y = y 2. Example 3. Consider the PDE u xx + u xy 2u yy + 1 = 0. (a) Reduce the equation to canonical form and find its general solution. (b) Find the solution u(x, y) satisfying the conditions u(x, 0) = u y (x, 0) = x. Constant coefficient equation Au xx + Bu xy + Cu yy = 0 The equation Au xx + Bu xy + Cu yy = 0, (4) where A, B, C are constants, can be solved by transforming it to canonical form. (1) If the equation (4) is hyperbolic, then the transformations ξ = y + λ 1 x, η = y + λ 2 x, where λ 1,2 are the roots of the characteristic equation Aλ 2 + Bλ + C = 0, transform equation (4) into its canonical form u ξη = 0. It follows that the solution is u(x, y) = f(ξ) + g(η) = f(y + λ 1 x) + g(y + λ 2 x), where f and g are arbitrary functions. (2) If the equation (4) is parabolic, then the transformations ξ = x, η = y + λx, where λ is the root of the characteristic equation Aλ 2 + Bλ + C = 0, transform equation (4) into its canonical form u ξξ = 0. It follows that the solution is u(x, y) = ξf(η) + g(η) = xf(y + λx) + g(y + λx), where f and g are arbitrary functions. 7

8 (3) If the equation (4) is elliptic, then the transformations ξ = y + λ 1 x, η = y + λ 2 x, where λ 1,2 are the roots of the characteristic equation transform equation (4) into its canonical form It follows that the solution is where f and g are arbitrary functions. substitution. Aλ 2 + Bλ + C = 0, u ξξ + u ηη = 0. u(x, y) = f(ξ) + g(η) = f(y + λ 1 x) + g(y + λ 2 x), Example 1. Find the general solution of the partial differential equation The proof of this form can be achieved by direct u xx + 4u xy + 4u yy = 0. Example 2. Find the general solution of the wave equation u tt = c 2 u xx, c > 0. Example 3. Find the general solution of the equation u xx + 9u yy = 0. Example 4. Find the general solution of the equation Exponential-type solutions u xx u xy 2u yy = sin y. By comparison with ordinary differential equations, we may look for solutions of Au xx + Bu xy + Cu yy + Du x + Eu y + F u = 0, where A, B, C, D, E, F are constants, in the form Substituting into the equation one get u(x, y) = e αx+βy. Aα 2 + Bαβ + Cβ 2 + Dα + Eβ + F = 0. Solving for β in terms of α, one may obtain a particular solution u(x, y) = k 1 e αx+β 1(α)y + k 2 e αx+β 2(α)y, where k 1 and k 2 are arbitrary constants. The particular solution can be used to guess the general solution. 8

9 Example 1. Find an exponential-type solution of the PDE Then find the general solution. u xx u yy 2u x + u = 0. Example 2. Find an exponential-type solution of the PDE and use it to find the general solution. u xx 2u xy + u yy 2u y + 2u x + u = 0 Example 3. Find a solution u(x, y) = e αx+β of the PDE and use it to find the general solution. u xx + u yy 2u y + u = 0 Remark. In some cases the exponential-type solutions may produce a set of useful particular solutions, but fail to suggest a general solution. Example. Determine a solution for the PDE u xx + u yy + 4u = 0 by letting u(x, y) = e αx+β. Can you propose a general solution? Exercises (1) Use direct integration to find the general solution of each of the following the PDEs: (a) xu xy + 2u y = y 2. (b) u yy = x 2 + y 2. (c) u yy xu y = x 2. (d) u xy = x y + 1. (e) xu xx = u x + x 2 y 2. (f) y 2 u yy + 2yu y = 1. (g) yu xy + u x = cos(x + y) y sin(x + y). (h) xu xy + 2u y = sin x x. (2) Use methods of ODE to find the general solution of each of the following PDEs: (a) xu xy + 3u y = y 3. 9

10 (b) u yy 4u y + 3u = 0. (c) x 2 u xx + 5xu x + 4u = 0. (3) Find a solution of the PDE satisfying the conditions u xy = x 2 y u y (0, y) = y 2, u(x, 1) = cos x. (4) Solve the following boundary value problems (a) u xy = 0, u x (x, 0) = cos x, u( π, y) = sin y. 2 (b) u xx = cos x, u(0, y) = y 2, u(π, y) = π sin y. (c) u yx = (1 + x 2 )y 2, u y (0, y) = y 3, u(x, 2) = cos x. (5) Classify the following equations as hyperbolic, parabolic, or elliptic and solve them. (a) u xx + 2u xy 3u yy = 0. (b) u xx 2u xy + u yy = 0. (c) u xx + au yy = 0, a > 0. (d) u xx 2u xy + 2u yy = 0. (5) Reduce each of following equations to its canonical form and obtain the general solution. (a) y 2 u xx 2yu xy + u yy = u x + 6y. (b) xu xx (x + y)u xy + yu yy = x + y x y (u x u y ). (c) u xx + au yy = 0, a > 0. (d) u xx 2u xy + 2u yy = 0. (6) Find an exponential-type solution of the following PDEs and use it to find the general solution. (a) u xx + 2u xy + u yy + u y + u x = 0. (b) 4u xx u yy 2u x + 4u y = 0. (c) 4u xx u yy 4u x + 2u y = 0. (d) 2u xx 2u xy + u yy + 4u = 0. (e) u xx 2u xy + u y u = 0. (7) Find the general solution for each of the following equations: (a) u xx 2u xy + 3u yy = e x. (b) u xx u xy 2u yy = sin y. (c) 4u xx 8u xy + 4u yy = 1. (d) u xx 2u x u y = x + y. (e) u xx + 4u xy + u yy + u x + u y = 3e 2x + xy. (e) u xx + u yy = x + ye y. 10

11 (8) Use the transformation w = (x y)u to simplify the Euler-Poisson-Darboux equation and solver it. (x y)u xy u x + u y = 0 (9) (a) Show that the transformations ξ = ln x, η = ln y transform the equation Ax 2 u xx + Bxyu xy + Cy 2 u yy + Dxu x + Eyu y + F u = G(x, y), where A, B, C, D, E, F are constants, into an equation with constant coefficients. (b) Use part (a) to solve the following equations (i) x 2 u xx y 2 u yy = xy. (ii) x 2 u xx xyu xy xu x = d Alembert s Solution of the Wave Equation The homogeneous wave equation in one (spatial) dimension has the form u tt c 2 u xx = 0, a < x < b, t > 0, (5) where c R is called the wave speed. Equation (5) is hyperbolic equation. The transformations ξ = x + ct, η = x ct, transform equation (5) into its canonical form u ξη = 0. It follows that the general solution of (5) is given by u(x, y) = φ(x + ct) + ψ(x ct), where φ and ψ are arbitrary functions. For a fixed t 0 > 0, the graph of the function ψ(x ct 0 ) has the same shape as the graph of the function ψ(x), except that it is shifted to the right by a distance ct 0. Therefore, the function ψ(x ct) represents a wave moving to the right with velocity c, and it is called a forward wave. The function φ(x + ct) is a wave traveling to the left with the same speed, and it is called a backward wave. Indeed c can be called the wave speed. Thus, any solution of the wave equation is the sum of two such traveling waves. Let us further discuss the general solution (5). Consider the (x, t) plane. The following two families of lines x ct = constant, x + ct = constant, are called the characteristics of the wave equation. For the wave equation, the characteristics are straight lines in the (x, t) plane with slopes ±1/c. 11

12 The Cauchy problem and d Alembert s formula The Cauchy problem for the one-dimensional homogeneous wave equation is given by u tt c 2 u xx = 0, < x <, t > 0, u(x, 0) = f(x), u t (x, 0) = g(x), < x <, (6) where f and g are given functions. Recall that the general solution of the wave equation is of the form u(x, y) = φ(x + ct) + ψ(x ct), where φ and ψ are arbitrary functions. Our aim is to find φ and ψ such that the initial conditions u(x, 0) = f(x), u t (x, 0) = g(x), < x <, are satisfied. The first condition u(x, 0) = f(x) gives The second condition u t (x, 0) = h 2 (x) gives φ(x) + ψ(x) = f(x). (7) cφ (x) cψ (x) = g(x). (8) To solve the two equations (7) and (8) we differentiate equation (7) to obtain Multiplying (9) by c and adding to equation (8) we obtain Therefore, φ (x) + ψ (x) = f (x). (9) 2cφ (x) = g(x) + cf (x). (10) φ(x) = 1 2 f(x) + 1 2c x x 0 g(z)dz + k, where k is a constant and x ct < x 0 < x + xt. Now equation (7) gives ψ(x) = 1 2 φ(x) 1 2c x x 0 g(z)dz k. Substituting these two expressions for f and g back into the solution formula we find u(x, t) = φ(x + ct) + ψ(x ct), u(t, x) = 1 2 [f(x + ct) + f(x ct)] + 1 2c x+ct x ct g(z)dz. In this manner, we have arrived at d Alembert s solution to the initial value problem for the wave equation on the entire line. 12

13 Theorem 6. (d Alembert s solution) The solution to the initial value problem is given by Remark. In order that the formula u tt c 2 u xx = 0, < x <, t > 0, u(x, 0) = f(x), u t (x, 0) = g(x), < x <, u(t, x) = 1 2 [f(x + ct) + f(x ct)] + 1 2c u(t, x) = 1 2 [f(x + ct) + f(x ct)] + 1 2c x+ct x ct x+ct x ct g(z)dz. g(z)dz, define a classical solution to the wave equation, f and g must be continuous. However, the formula itself makes sense for more general initial conditions. We will continue to treat the resulting functions as solutions as they do fit under the more general concept of weak solution. Example 1. Use d Alembert s formula to find the solution of the Cauchy problem u tt u xx = 0, < x <, t > 0, u(x, 0) = sin x, u t (x, 0) = x 2, < x <. Example 2. Use d Alembert s formula to find the solution of the Cauchy problem Example 3. Consider the Cauchy problem u tt 4u xx = 0, < x <, t > 0, u(x, 0) = e x, u t (x, 0) = 0, < x <. u tt 9u xx = 0, < x <, t > 0, u(x, 0) = f(x), u t (x, 0) = g(x), < x <, where f(x) = g(x) = { 1, x 2; 0, x > 2. (a) Find u(x, t). (b) Find u(0, 1 3 ). 13

14 Exercises Use d Alembert s formula to find the solution of the following Cauchy problems (a) u tt u xx = 0, < x <, t > 0, u(x, 0) = x 2, u t(x, 0) = 0, < x <. (b) u tt u xx = 0, < x <, t > 0, u(x, 0) = x, u t (x, 0) = 1, < x <. (c) u tt 16u xx = 0, < x <, t > 0, u(x, 0) = ln(1 + x 2 ), u t (x, 0) = 2, < x <. (d) u tt 4u xx = 0, < x <, t > 0, u(x, 0) = sin x, u t (x, 0) = cos x, < x <. (e) u tt u xx = 0, < x <, t > 0, u(x, 0) = x 2, u t(x, 0) = 0, < x <. (f) u tt u xx = 0, < x <, t > 0, u(x, 0) = 0, u t (x, 0) = sinh(2x), < x <. (g) u tt 16u xx = 0, < x <, t > 0, 0, < x < 1; u(x, 0) = 1, 1 x 1;, u t (x, 0) = 0, < x <. 0, x > 1, (h) u tt c 2 u xx = 0, < x <, t > 0, u(x, 0) = 0, u t (x, 0) = e x, < x <. 3.4 Separation of variables Separation of variables is a useful method for solving partial differential equations. In this method we look for solutions in the form of a product of functions of a single independent variable: Example 1. Find a solution of the equation using the separation of variables method. Example 2. Find solutions of the equation where a is a constant, in the form u(x 1, x 2, ) = f 1 (x 1 )f 2 (x 2 ). u t = 4u xx u t = a 2 (u xx + u yy ), u(t, x, y) = T (t)x(x)y (y). 14

15 Remarks. (1) In most cases a bounded solution will be required. The separations constants must be selected to satisfy this requirement. (2) The method of separation of variables is valuable for solving a number of important problems in mathematical physics, yet it fails for many partial differential equations and boundary value problems. Type of the equation and forms of the boundary conditions are all important for the success of the method. Example. Determine if the equation (x + y)u xx + u yy = 0 has a solution in the form or not. u(x, y) = X(x)Y (y) Exercises (1) Determine whether the following equations can be solved be the method of separation of variables. If the equation is solvable by the method of separation of variables, then solve the equation. (a) u xy u = 0. (b) u xy = 0. (c) u xx u yy 2u y = 0. (d) u xx u yy + 2u x 2u y + u = 0. (e) u xx u xy + u yy = 2x. (f) u tt u xx = 0. (g) u t = u xx. (h) u xx u xy + u yy = 0. (i) t 2 u tt x 2 u xx = 0. (2) Find a solution for the following boundary value problems (a) u tt u xx = 0, u(x, 0) = u(0, t) = 0. (b) u xx u yy 2u y = 0, u x (0, y) = u(x, 0) = 0. (c) u t = u xx, u x (0, t) = 0. 15

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